# How do you solve x^2+4x+7=0?

Aug 12, 2015

Try factoring first. It won't work, so either use the quadratic formula or complete the square, as you prefer.

#### Explanation:

${x}^{2} + 4 x + 7 = 0$

The expression will not factor using integers.

Use your favorite of either the quadratic formula or completing the square.

Completing the square

Because the leading coefficient is $1$, and the coefficient of $x$ is even, completing the square is a good choice for this equation.

${x}^{2} + 4 x \text{ " " } = - 7$

${x}^{2} + 4 x + 4 = - 7 + 4$

${\left(x + 2\right)}^{2} = - 3$

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

$x + 2 = \pm \sqrt{- 3} = \pm i \sqrt{3}$

$x = - 2 \pm i \sqrt{3}$

${x}^{2} + 4 x + 7 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(4\right) \pm \sqrt{{\left(4\right)}^{2} - 4 \left(1\right) \left(7\right)}}{2 \left(1\right)}$

$= \frac{- 4 \pm \sqrt{16 - 28}}{2}$

$= \frac{- 4 \pm \sqrt{- 12}}{2}$

Which has no solution in the Real Numbers, but has 2 imaginary solutions in the Complex Numbers.

$= \frac{- 4 \pm 2 i \sqrt{3}}{2}$

$= - 2 \pm i \sqrt{3}$